如何在 Haskell 中获取无限列表的每个第 N 个元素? [英] How to get every Nth element of an infinite list in Haskell?

问题描述

更具体地说，如何从现有的无限列表中生成每个第 N 个元素的新列表?

More specifically, how do I generate a new list of every Nth element from an existing infinite list?

例如如果列表是 [5, 3, 0, 1, 8, 0, 3, 4, 0, 93, 211, 0 ...] 然后获取每个第三个元素将导致此列表 <代码>[0,0,0,0,0 ...]

E.g. if the list is [5, 3, 0, 1, 8, 0, 3, 4, 0, 93, 211, 0 ...] then getting every 3rd element would result in this list [0,0,0,0,0 ...]

推荐答案

我使用 drop 的版本:

My version using drop:

every n xs = case drop (n-1) xs of
              y : ys -> y : every n ys
              [] -> []

这也适用于有限列表.仅对于无限列表，Charles Stewart 的解决方案略短.

this also works for finite lists. For infinite lists only, Charles Stewart's solution is slightly shorter.

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